Problem 4

Let

R

and

S

be different points on a circle

\Omega

such that

RS

is not a diameter. Let

\ell

be the tangent line to

\Omega

R

. Point

T

is such that

S

is the midpoint of

RT

. Point

J

is chosen on the shorter arc

RS

\Omega

so that the circumcircle

\Gamma

of triangle

JST

intersects

\ell

at two distinct points. Let

A

be the common point of

\Gamma

and

\ell

that is closer to

R

. Line

AJ

meets

\Omega

again at

K

. Prove that line

KT

is tangent to

\Gamma

Step 1 of 6: Introduce the reflected point

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B=2S-A,\qquad ARBT\text{ is a parallelogram}

Detailed analysis

Let B be the reflection of A in S. Since S is the midpoint of both RT and AB, the quadrilateral with vertices A, R, B, T in that order is a parallelogram. In particular, AT is parallel to RB, and A, S, B are collinear.